S1 June 2014 (R) Q5

EdexcelOld spec12 marksData Processing

5. The table shows the time, to the nearest minute, spent waiting for a taxi by each of 80 people one Sunday afternoon.

Waiting time (in minutes)	Frequency
2–4	15
5–6	9
7	6
8	24
9–10	14
11–15	12

(a) Write down the upper class boundary for the 2–4 minute interval. (1)

A histogram is drawn to represent these data. The height of the tallest bar is 6 cm.

(b) Calculate the height of the second tallest bar. (3)

(d) Use linear interpolation to estimate the median, the lower quartile and the upper quartile of the waiting times. (4)

(e) Describe the skewness of these data, giving a reason for your answer. (2)

Scheme	Marks
4.5	B1
	(1)

Mark scheme (b)

Scheme

Marks

Time	Frequency density
2-4	5
5-6	4.5
7	6
8	24
9-10	7
11-15	2.4

M1
A1

f.d = 24 is represented as 6cm, so
f.d. = 7 is represented as 1.75(cm)

(3)

Notes

M1 for evidence of f/w (at least 3 f.d. found). May be implied by a correct answer.

A1 for identifying 9-10 as 2nd highest bar from correct working e.g. \(24x = 6 \times 7\)

A1 for 1.75(cm). Correct answer only 3/3

Mark scheme (c)

Scheme	Marks
\(\dfrac{1}{3} \times 15 + 9 + \dfrac{1}{2} \times 6, = 17\)	M1, A1
	(2)

Notes

M1 for a correct expression. May interpolate e.g. \(\left[24 + \frac{1}{2} \times 6 - \frac{2}{3} \times 15\right]\) or (27 – 10)

A1 for 17

Mark scheme (d)

Scheme	Marks
\(\text{Median} = 7.5 + \dfrac{40 - 30}{24} \times 1 = 7.91666\ldots\) awrt 7.92 or 7.93(75)	M1 A1
\(Q_1 = 4.5 + \dfrac{20 - 15}{9} \times 2 = 5.6111111\ldots\) awrt 5.61 or 5.66(666…)	A1
\(Q_3 = 8.5 + \dfrac{60 - 54}{14} \times 2 = 9.357142\ldots\) awrt 9.36 or 9.46(4285…)	A1
	(4)

Notes

M1 for one correct fraction in an expression for \(Q_1\), \(Q_2\) or \(Q_3\)

1st A1 for \(Q_2\) awrt 7.92 (or 7.94 if (\(n\) +1) used – look for 40.5 instead of 40)

2nd A1 for \(Q_1\) awrt 5.61 (or 5.67 if (\(n\) +1) used – look for 20.25 instead of 20)

3rd A1 for \(Q_3\) awrt 9.36 (or 9.46 if (\(n\) +1) used – look for 60.75 instead of 60)

NB watch out for working down e.g. \(8.5 - \dfrac{14}{24} \times 1\) for \(Q_2\)

Mark scheme (e)

Scheme	Marks
\(Q_3 - Q_2\ (= 1.4 \text{ or } 1.5) \lt Q_2 - Q_1\ (= 2.3)\) or (Mean) < Median < Mode	B1ft
Therefore negative skew	dB1cao
	(2)
	(12 marks)

Notes

1st B1ft for a correct comparison of their quartiles e.g. \(Q_2\) closer to \(Q_3\) or using at least two of Mean < Median < Mode (must state mean or mode if using this method).
N.B. Mean = 7.71875, mode = 8

2nd B1cao dependent on 1st B1 being awarded for negative skew only (no ft)